(1)圆心C(m,0),(-1<m<1),
则⊙C的半径为:r=
从而⊙C的方程为(x-m)2+y2=1-m2,
椭圆D的标准方程为:
(2)当b=1时,椭圆D的方程为
x221+y2=1,
设椭圆D上任意一点S(x1,y1),
x122+y12=1,y12=1-
∵SC2=(x1-m) 2+y12
=(x1-m) 2+1-
= (x1-2m)2+1-m2
≥1-m2=r2,
所以SC≥r.
从而椭圆D上的任意一点都不存在⊙C的内部.
(3)
OL=b2+1为定值.
证明:设点P(x1,y1),Q(x2,y2),
则由题意,得N(x1,-y1),x1≠x2,y1≠±y2,
从而直线PQ的方程为(y2-y1)x-(x2-x1)y+x2y1-x1y2=0,
令y=0,得xM=
x1y2-x2y1 y2-y1,
∵直线QN的方程为(y2+y1)x-(x2-x1)y-x1y2-x2y1=0,
令y=0,得xL=
x2y1+x1y2y2+y1.
∵点P,Q在椭圆D上,
∴
x12b2+1+
y12b2=1,
x22b2+1+
y22b2=1,
∴x12=b2+1-
b2+1b2y12,x22=b2+1-
b2+1b2y22,
∴xM•xL=
(b2+1-
b2+1b2y12)y22-(b2+1-
b2+1b2y22 )y12y22-y12
=
(b2+1)(y22-y12)y22-y12=b2+1.
∴
OM•
OL=xM•xL=b2+1为定值.